Answer :
Answer: mark me as brainlist please
Explanation:
The given C code defines a structure `marks` with two members `p` and `c`, each defined as 3-bit integers (`int p:3;` and `int c:3;`).
Let's analyze the structure definition and initialization:
- `int p:3;` implies `p` can hold values in the range -4 to 3 (since it's a 3-bit signed integer).
- `int c:3;` similarly implies `c` can also hold values in the range -4 to 3.
Now, the structure `marks` is initialized with `{2, -6}`:
- `s.p = 2;` This assigns the value 2 to `s.p`. In 3-bit signed integer representation, 2 is represented as `010` in binary (considering the sign bit).
- `s.c = -6;` This assigns the value -6 to `s.c`. In 3-bit signed integer representation, -6 is represented as `110` in binary (again, considering the sign bit).
When we print `s.p` and `s.c` using `printf("%d %d", s.p, s.c);`, the output will be:
- `s.p` is 2.
- `s.c` in 3-bit signed integer representation is `-2` (because `110` in 3-bit signed integer format corresponds to -2 in decimal).
Therefore, the correct output of the program will be:
```
2 -2
```
So, the answer is not provided exactly in the given options, but the closest option would be:
(b) 2 1
Answer:
The given C code snippet defines a structure marks with two members p and c, both of which are bit-fields of size 3 (meaning they can hold values from -4 to 3, inclusive).Let's analyze the code step by step:struct marks {
int p:3; // Bit-field for 'p' with 3 bits
int c:3; // Bit-field for 'c' with 3 bits
};
void main() {
struct marks s = {2, -6}; // Initializing struct marks s with p = 2 and c = -6
printf("%d %d", s.p, s.c); // Printing values of s.p and s.c
}Explanation:Struct Definition:struct marks is defined with two members p and c, each being a 3-bit signed integer bit-field.Initialization:struct marks s = {2, -6}; initializes s.p with 2 and s.c with -6.Printing:printf("%d %d", s.p, s.c); prints the values of s.p and s.c.Evaluation of Output:s.p is initialized with 2, which is within the range of a 3-bit signed integer (-4 to 3). So, s.p remains 2.s.c is initialized with -6, which is outside the range of a 3-bit signed integer. When initializing a bit-field with a value outside its range, the behavior is implementation-defined according to the C standard (ISO/IEC 9899:1999). Common implementations might reinterpret -6 in a way that fits within the bit-field's range (-6 % 8 = 2 in this case, because of two's complement representation).Conclusion:Based on the analysis:s.p will be 2.s.c will likely be interpreted as 2 due to the implementation-defined behavior.Therefore, the output of the code will be:2 2So, the correct answer is (c) 2 2.